如何判断字符串是否在Python中重复?

问题:如何判断字符串是否在Python中重复?

我正在寻找一种方法来测试给定的字符串是否对整个字符串重复。

例子:

[
    '0045662100456621004566210045662100456621',             # '00456621'
    '0072992700729927007299270072992700729927',             # '00729927'
    '001443001443001443001443001443001443001443',           # '001443'
    '037037037037037037037037037037037037037037037',        # '037'
    '047619047619047619047619047619047619047619',           # '047619'
    '002457002457002457002457002457002457002457',           # '002457'
    '001221001221001221001221001221001221001221',           # '001221'
    '001230012300123001230012300123001230012300123',        # '00123'
    '0013947001394700139470013947001394700139470013947',    # '0013947'
    '001001001001001001001001001001001001001001001001001',  # '001'
    '001406469760900140646976090014064697609',              # '0014064697609'
]

是重复的字符串,并且

[
    '004608294930875576036866359447',
    '00469483568075117370892018779342723',
    '004739336492890995260663507109',
    '001508295625942684766214177978883861236802413273',
    '007518796992481203',
    '0071942446043165467625899280575539568345323741',
    '0434782608695652173913',
    '0344827586206896551724137931',
    '002481389578163771712158808933',
    '002932551319648093841642228739',
    '0035587188612099644128113879',
    '003484320557491289198606271777',
    '00115074798619102416570771',
]

是那些没有的例子。

我得到的字符串的重复部分可能会很长,并且字符串本身可以是500个或更多字符,因此循环遍历每个字符以尝试构建模式,然后检查模式与字符串的其余部分似乎很慢。将其乘以可能的数百个字符串,就看不到任何直观的解决方案。

我对正则表达式进行了一些研究,当您知道要查找的内容时,或者至少在寻找所需模式的长度时,它们似乎非常有用。不幸的是,我都不知道。

我怎么知道一个字符串是否在重复本身,如果是,最短的重复子序列是什么?

I’m looking for a way to test whether or not a given string repeats itself for the entire string or not.

Examples:

[
    '0045662100456621004566210045662100456621',             # '00456621'
    '0072992700729927007299270072992700729927',             # '00729927'
    '001443001443001443001443001443001443001443',           # '001443'
    '037037037037037037037037037037037037037037037',        # '037'
    '047619047619047619047619047619047619047619',           # '047619'
    '002457002457002457002457002457002457002457',           # '002457'
    '001221001221001221001221001221001221001221',           # '001221'
    '001230012300123001230012300123001230012300123',        # '00123'
    '0013947001394700139470013947001394700139470013947',    # '0013947'
    '001001001001001001001001001001001001001001001001001',  # '001'
    '001406469760900140646976090014064697609',              # '0014064697609'
]

are strings which repeat themselves, and

[
    '004608294930875576036866359447',
    '00469483568075117370892018779342723',
    '004739336492890995260663507109',
    '001508295625942684766214177978883861236802413273',
    '007518796992481203',
    '0071942446043165467625899280575539568345323741',
    '0434782608695652173913',
    '0344827586206896551724137931',
    '002481389578163771712158808933',
    '002932551319648093841642228739',
    '0035587188612099644128113879',
    '003484320557491289198606271777',
    '00115074798619102416570771',
]

are examples of ones that do not.

The repeating sections of the strings I’m given can be quite long, and the strings themselves can be 500 or more characters, so looping through each character trying to build a pattern then checking the pattern vs the rest of the string seems awful slow. Multiply that by potentially hundreds of strings and I can’t see any intuitive solution.

I’ve looked into regexes a bit and they seem good for when you know what you’re looking for, or at least the length of the pattern you’re looking for. Unfortunately, I know neither.

How can I tell if a string is repeating itself and if it is, what the shortest repeating subsequence is?


回答 0

这是一个简洁的解决方案,它避免了正则表达式和缓慢的Python内循环:

def principal_period(s):
    i = (s+s).find(s, 1, -1)
    return None if i == -1 else s[:i]

有关基准测试结果,请参阅由@davidism发起的Community Wiki答案。综上所述,

David Zhang的解决方案无疑是赢家,对于大型示例集,其性能至少比其他同类产品高出5倍。

(那个答案的话,不是我的。)

这是基于这样的观察,即且仅当字符串等于其自身的非平凡旋转时,它才是周期性的。@AleksiTorhamo的荣誉,意识到我们可以从sin 的第一次出现的索引中恢复本金周期(s+s)[1:-1],并告知我Python的可选参数start和自end变量string.find

Here’s a concise solution which avoids regular expressions and slow in-Python loops:

def principal_period(s):
    i = (s+s).find(s, 1, -1)
    return None if i == -1 else s[:i]

See the Community Wiki answer started by @davidism for benchmark results. In summary,

David Zhang’s solution is the clear winner, outperforming all others by at least 5x for the large example set.

(That answer’s words, not mine.)

This is based on the observation that a string is periodic if and only if it is equal to a nontrivial rotation of itself. Kudos to @AleksiTorhamo for realizing that we can then recover the principal period from the index of the first occurrence of s in (s+s)[1:-1], and for informing me of the optional start and end arguments of Python’s string.find.


回答 1

这是使用正则表达式的解决方案。

import re

REPEATER = re.compile(r"(.+?)\1+$")

def repeated(s):
    match = REPEATER.match(s)
    return match.group(1) if match else None

迭代问题中的示例:

examples = [
    '0045662100456621004566210045662100456621',
    '0072992700729927007299270072992700729927',
    '001443001443001443001443001443001443001443',
    '037037037037037037037037037037037037037037037',
    '047619047619047619047619047619047619047619',
    '002457002457002457002457002457002457002457',
    '001221001221001221001221001221001221001221',
    '001230012300123001230012300123001230012300123',
    '0013947001394700139470013947001394700139470013947',
    '001001001001001001001001001001001001001001001001001',
    '001406469760900140646976090014064697609',
    '004608294930875576036866359447',
    '00469483568075117370892018779342723',
    '004739336492890995260663507109',
    '001508295625942684766214177978883861236802413273',
    '007518796992481203',
    '0071942446043165467625899280575539568345323741',
    '0434782608695652173913',
    '0344827586206896551724137931',
    '002481389578163771712158808933',
    '002932551319648093841642228739',
    '0035587188612099644128113879',
    '003484320557491289198606271777',
    '00115074798619102416570771',
]

for e in examples:
    sub = repeated(e)
    if sub:
        print("%r: %r" % (e, sub))
    else:
        print("%r does not repeat." % e)

…产生以下输出:

'0045662100456621004566210045662100456621': '00456621'
'0072992700729927007299270072992700729927': '00729927'
'001443001443001443001443001443001443001443': '001443'
'037037037037037037037037037037037037037037037': '037'
'047619047619047619047619047619047619047619': '047619'
'002457002457002457002457002457002457002457': '002457'
'001221001221001221001221001221001221001221': '001221'
'001230012300123001230012300123001230012300123': '00123'
'0013947001394700139470013947001394700139470013947': '0013947'
'001001001001001001001001001001001001001001001001001': '001'
'001406469760900140646976090014064697609': '0014064697609'
'004608294930875576036866359447' does not repeat.
'00469483568075117370892018779342723' does not repeat.
'004739336492890995260663507109' does not repeat.
'001508295625942684766214177978883861236802413273' does not repeat.
'007518796992481203' does not repeat.
'0071942446043165467625899280575539568345323741' does not repeat.
'0434782608695652173913' does not repeat.
'0344827586206896551724137931' does not repeat.
'002481389578163771712158808933' does not repeat.
'002932551319648093841642228739' does not repeat.
'0035587188612099644128113879' does not repeat.
'003484320557491289198606271777' does not repeat.
'00115074798619102416570771' does not repeat.

正则表达式(.+?)\1+$分为三个部分:

  1. (.+?)是一个匹配组,其中包含至少一个(但尽可能少)任何字符(因为+?不是贪婪)。

  2. \1+ 在第一部分中检查匹配组的至少一个重复。

  3. $检查字符串的结尾,以确保在重复的子字符串之后没有多余的,非重复的内容(并使用re.match()确保在重复的子字符串之前没有非重复的文本)。

在Python 3.4和更高版本中,您可以删除$re.fullmatch()来代替,或者(在任何Python中至少可以追溯到2.3)使用另一种方式并re.search()与regex一起使用^(.+?)\1+$,所有这些都比其他人更受个人喜好。

Here’s a solution using regular expressions.

import re

REPEATER = re.compile(r"(.+?)\1+$")

def repeated(s):
    match = REPEATER.match(s)
    return match.group(1) if match else None

Iterating over the examples in the question:

examples = [
    '0045662100456621004566210045662100456621',
    '0072992700729927007299270072992700729927',
    '001443001443001443001443001443001443001443',
    '037037037037037037037037037037037037037037037',
    '047619047619047619047619047619047619047619',
    '002457002457002457002457002457002457002457',
    '001221001221001221001221001221001221001221',
    '001230012300123001230012300123001230012300123',
    '0013947001394700139470013947001394700139470013947',
    '001001001001001001001001001001001001001001001001001',
    '001406469760900140646976090014064697609',
    '004608294930875576036866359447',
    '00469483568075117370892018779342723',
    '004739336492890995260663507109',
    '001508295625942684766214177978883861236802413273',
    '007518796992481203',
    '0071942446043165467625899280575539568345323741',
    '0434782608695652173913',
    '0344827586206896551724137931',
    '002481389578163771712158808933',
    '002932551319648093841642228739',
    '0035587188612099644128113879',
    '003484320557491289198606271777',
    '00115074798619102416570771',
]

for e in examples:
    sub = repeated(e)
    if sub:
        print("%r: %r" % (e, sub))
    else:
        print("%r does not repeat." % e)

… produces this output:

'0045662100456621004566210045662100456621': '00456621'
'0072992700729927007299270072992700729927': '00729927'
'001443001443001443001443001443001443001443': '001443'
'037037037037037037037037037037037037037037037': '037'
'047619047619047619047619047619047619047619': '047619'
'002457002457002457002457002457002457002457': '002457'
'001221001221001221001221001221001221001221': '001221'
'001230012300123001230012300123001230012300123': '00123'
'0013947001394700139470013947001394700139470013947': '0013947'
'001001001001001001001001001001001001001001001001001': '001'
'001406469760900140646976090014064697609': '0014064697609'
'004608294930875576036866359447' does not repeat.
'00469483568075117370892018779342723' does not repeat.
'004739336492890995260663507109' does not repeat.
'001508295625942684766214177978883861236802413273' does not repeat.
'007518796992481203' does not repeat.
'0071942446043165467625899280575539568345323741' does not repeat.
'0434782608695652173913' does not repeat.
'0344827586206896551724137931' does not repeat.
'002481389578163771712158808933' does not repeat.
'002932551319648093841642228739' does not repeat.
'0035587188612099644128113879' does not repeat.
'003484320557491289198606271777' does not repeat.
'00115074798619102416570771' does not repeat.

The regular expression (.+?)\1+$ is divided into three parts:

  1. (.+?) is a matching group containing at least one (but as few as possible) of any character (because +? is non-greedy).

  2. \1+ checks for at least one repetition of the matching group in the first part.

  3. $ checks for the end of the string, to ensure that there’s no extra, non-repeating content after the repeated substrings (and using re.match() ensures that there’s no non-repeating text before the repeated substrings).

In Python 3.4 and later, you could drop the $ and use re.fullmatch() instead, or (in any Python at least as far back as 2.3) go the other way and use re.search() with the regex ^(.+?)\1+$, all of which are more down to personal taste than anything else.


回答 2

您可以观察到对于要考虑重复的字符串,必须将其长度除以重复序列的长度。鉴于此,这是一个生成长度为从1n / 2包括的长度的除数的解决方案,将原始字符串分成具有除数长度的子字符串,并测试结果集的相等性:

from math import sqrt, floor

def divquot(n):
    if n > 1:
        yield 1, n
    swapped = []
    for d in range(2, int(floor(sqrt(n))) + 1):
        q, r = divmod(n, d)
        if r == 0:
            yield d, q
            swapped.append((q, d))
    while swapped:
        yield swapped.pop()

def repeats(s):
    n = len(s)
    for d, q in divquot(n):
        sl = s[0:d]
        if sl * q == s:
            return sl
    return None

编辑:在Python 3中,/运算符已更改为默认情况下进行浮点除法。要从intPython 2中进行除法,可以改用//运算符。感谢@ TigerhawkT3引起我的注意。

//运算符在Python 2和Python 3中都执行整数除法,因此我更新了答案以支持这两个版本。现在,我们测试以查看所有子串是否相等的部分是使用all和生成器表达式的短路操作。

更新:为响应原始问题的更改,现在对代码进行了更新,以返回最小的重复子字符串(如果存在),None如果不存在,则返回最小。@godlygeek建议使用divmod减少divisors生成器上的迭代次数,并且代码也已更新为与此匹配。现在,它n以升序返回所有正数除数,不包括n其本身。

进一步更新以提高性能:经过多次测试,我得出的结论是,简单地测试字符串相等性具有Python中任何切片或迭代器解决方案中最好的性能。因此,我从@ TigerhawkT3的书中抽出了叶子,并更新了我的解决方案。现在它的速度是以前的6倍,比Tigerhawk的解决方案快得多,但比David的解决方案慢。

You can make the observation that for a string to be considered repeating, its length must be divisible by the length of its repeated sequence. Given that, here is a solution that generates divisors of the length from 1 to n / 2 inclusive, divides the original string into substrings with the length of the divisors, and tests the equality of the result set:

from math import sqrt, floor

def divquot(n):
    if n > 1:
        yield 1, n
    swapped = []
    for d in range(2, int(floor(sqrt(n))) + 1):
        q, r = divmod(n, d)
        if r == 0:
            yield d, q
            swapped.append((q, d))
    while swapped:
        yield swapped.pop()

def repeats(s):
    n = len(s)
    for d, q in divquot(n):
        sl = s[0:d]
        if sl * q == s:
            return sl
    return None

EDIT: In Python 3, the / operator has changed to do float division by default. To get the int division from Python 2, you can use the // operator instead. Thank you to @TigerhawkT3 for bringing this to my attention.

The // operator performs integer division in both Python 2 and Python 3, so I’ve updated the answer to support both versions. The part where we test to see if all the substrings are equal is now a short-circuiting operation using all and a generator expression.

UPDATE: In response to a change in the original question, the code has now been updated to return the smallest repeating substring if it exists and None if it does not. @godlygeek has suggested using divmod to reduce the number of iterations on the divisors generator, and the code has been updated to match that as well. It now returns all positive divisors of n in ascending order, exclusive of n itself.

Further update for high performance: After multiple tests, I’ve come to the conclusion that simply testing for string equality has the best performance out of any slicing or iterator solution in Python. Thus, I’ve taken a leaf out of @TigerhawkT3 ‘s book and updated my solution. It’s now over 6x as fast as before, noticably faster than Tigerhawk’s solution but slower than David’s.


回答 3

以下是针对此问题的各种答案的一些基准。有一些令人惊讶的结果,包括不同的性能,具体取决于所测试的字符串。

修改了某些功能以使其与Python 3兼容(主要是通过替换///以确保整数除法)。如果发现错误,请添加功能或添加另一个测试字符串,请在Python聊天室中 ping @ZeroPiraeus 。

总结:对于此处由OP提供的大量示例数据,最佳和最差解决方案之间存在大约50倍的差异(通过评论)。David Zhang的解决方案无疑是赢家,对于大型示例集,其解决方案比其他所有解决方案都高出约5倍。

在极大的“不匹配”情况下,几个答案非常慢。否则,根据测试,功能似乎是相等的,或者是明显的赢家。

以下是结果,包括使用matplotlib和seaborn绘制的图以显示不同的分布:


语料库1(提供的示例-小集)

mean performance:
 0.0003  david_zhang
 0.0009  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0015  carpetpython
 0.0029  tigerhawk_1
 0.0031  davidism
 0.0035  saksham
 0.0046  shashank
 0.0052  riad
 0.0056  piotr

median performance:
 0.0003  david_zhang
 0.0008  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0014  carpetpython
 0.0027  tigerhawk_1
 0.0031  davidism
 0.0038  saksham
 0.0044  shashank
 0.0054  riad
 0.0058  piotr


语料库2(提供的示例-大集合)

mean performance:
 0.0006  david_zhang
 0.0036  tigerhawk_2
 0.0036  antti
 0.0037  zero
 0.0039  carpetpython
 0.0052  shashank
 0.0056  piotr
 0.0066  davidism
 0.0120  tigerhawk_1
 0.0177  riad
 0.0283  saksham

median performance:
 0.0004  david_zhang
 0.0018  zero
 0.0022  tigerhawk_2
 0.0022  antti
 0.0024  carpetpython
 0.0043  davidism
 0.0049  shashank
 0.0055  piotr
 0.0061  tigerhawk_1
 0.0077  riad
 0.0109  saksham


语料库3(边缘案例)

mean performance:
 0.0123  shashank
 0.0375  david_zhang
 0.0376  piotr
 0.0394  carpetpython
 0.0479  antti
 0.0488  tigerhawk_2
 0.2269  tigerhawk_1
 0.2336  davidism
 0.7239  saksham
 3.6265  zero
 6.0111  riad

median performance:
 0.0107  tigerhawk_2
 0.0108  antti
 0.0109  carpetpython
 0.0135  david_zhang
 0.0137  tigerhawk_1
 0.0150  shashank
 0.0229  saksham
 0.0255  piotr
 0.0721  davidism
 0.1080  zero
 1.8539  riad


在此处获得测试和原始结果。

Here are some benchmarks for the various answers to this question. There were some surprising results, including wildly different performance depending on the string being tested.

Some functions were modified to work with Python 3 (mainly by replacing / with // to ensure integer division). If you see something wrong, want to add your function, or want to add another test string, ping @ZeroPiraeus in the Python chatroom.

In summary: there’s about a 50x difference between the best- and worst-performing solutions for the large set of example data supplied by OP here (via this comment). David Zhang’s solution is the clear winner, outperforming all others by around 5x for the large example set.

A couple of the answers are very slow in extremely large “no match” cases. Otherwise, the functions seem to be equally matched or clear winners depending on the test.

Here are the results, including plots made using matplotlib and seaborn to show the different distributions:


Corpus 1 (supplied examples – small set)

mean performance:
 0.0003  david_zhang
 0.0009  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0015  carpetpython
 0.0029  tigerhawk_1
 0.0031  davidism
 0.0035  saksham
 0.0046  shashank
 0.0052  riad
 0.0056  piotr

median performance:
 0.0003  david_zhang
 0.0008  zero
 0.0013  antti
 0.0013  tigerhawk_2
 0.0014  carpetpython
 0.0027  tigerhawk_1
 0.0031  davidism
 0.0038  saksham
 0.0044  shashank
 0.0054  riad
 0.0058  piotr


Corpus 2 (supplied examples – large set)

mean performance:
 0.0006  david_zhang
 0.0036  tigerhawk_2
 0.0036  antti
 0.0037  zero
 0.0039  carpetpython
 0.0052  shashank
 0.0056  piotr
 0.0066  davidism
 0.0120  tigerhawk_1
 0.0177  riad
 0.0283  saksham

median performance:
 0.0004  david_zhang
 0.0018  zero
 0.0022  tigerhawk_2
 0.0022  antti
 0.0024  carpetpython
 0.0043  davidism
 0.0049  shashank
 0.0055  piotr
 0.0061  tigerhawk_1
 0.0077  riad
 0.0109  saksham


Corpus 3 (edge cases)

mean performance:
 0.0123  shashank
 0.0375  david_zhang
 0.0376  piotr
 0.0394  carpetpython
 0.0479  antti
 0.0488  tigerhawk_2
 0.2269  tigerhawk_1
 0.2336  davidism
 0.7239  saksham
 3.6265  zero
 6.0111  riad

median performance:
 0.0107  tigerhawk_2
 0.0108  antti
 0.0109  carpetpython
 0.0135  david_zhang
 0.0137  tigerhawk_1
 0.0150  shashank
 0.0229  saksham
 0.0255  piotr
 0.0721  davidism
 0.1080  zero
 1.8539  riad


The tests and raw results are available here.


回答 4

非正则表达式解决方案:

def repeat(string):
    for i in range(1, len(string)//2+1):
        if not len(string)%len(string[0:i]) and string[0:i]*(len(string)//len(string[0:i])) == string:
            return string[0:i]

更快的非正则表达式解决方案,这要感谢@ThatWeirdo(请参见评论):

def repeat(string):
    l = len(string)
    for i in range(1, len(string)//2+1):
        if l%i: continue
        s = string[0:i]
        if s*(l//i) == string:
            return s

上面的解决方案很少比原始解决方案慢几个百分点,但通常会快很多-有时快很多。对于较长的字符串,它仍然不比davidism的快,而对于短字符串,zero的正则表达式解决方案更胜一筹。它以大约1000-1500个字符的字符串显示出来,速度最快(根据github上davidism的测试-请参见他的回答)。无论如何,在我测试的所有情况下,它都是第二快的(或更好的)。谢谢,ThatWeirdo。

测试:

print(repeat('009009009'))
print(repeat('254725472547'))
print(repeat('abcdeabcdeabcdeabcde'))
print(repeat('abcdefg'))
print(repeat('09099099909999'))
print(repeat('02589675192'))

结果:

009
2547
abcde
None
None
None

Non-regex solution:

def repeat(string):
    for i in range(1, len(string)//2+1):
        if not len(string)%len(string[0:i]) and string[0:i]*(len(string)//len(string[0:i])) == string:
            return string[0:i]

Faster non-regex solution, thanks to @ThatWeirdo (see comments):

def repeat(string):
    l = len(string)
    for i in range(1, len(string)//2+1):
        if l%i: continue
        s = string[0:i]
        if s*(l//i) == string:
            return s

The above solution is very rarely slower than the original by a few percent, but it’s usually a good bit faster – sometimes a whole lot faster. It’s still not faster than davidism’s for longer strings, and zero’s regex solution is superior for short strings. It comes out to the fastest (according to davidism’s test on github – see his answer) with strings of about 1000-1500 characters. Regardless, it’s reliably second-fastest (or better) in all cases I tested. Thanks, ThatWeirdo.

Test:

print(repeat('009009009'))
print(repeat('254725472547'))
print(repeat('abcdeabcdeabcdeabcde'))
print(repeat('abcdefg'))
print(repeat('09099099909999'))
print(repeat('02589675192'))

Results:

009
2547
abcde
None
None
None

回答 5

首先,将字符串减半,只要它是“ 2部分”重复项即可。如果重复数为偶数,则会减少搜索空间。然后,继续寻找最小的重复字符串,检查是否通过将越来越大的子字符串拆分成完整的字符串而只得到空值。仅length // 2需要测试最多的子字符串,因为任何重复的内容都不会重复。

def shortest_repeat(orig_value):
    if not orig_value:
        return None

    value = orig_value

    while True:
        len_half = len(value) // 2
        first_half = value[:len_half]

        if first_half != value[len_half:]:
            break

        value = first_half

    len_value = len(value)
    split = value.split

    for i in (i for i in range(1, len_value // 2) if len_value % i == 0):
        if not any(split(value[:i])):
            return value[:i]

    return value if value != orig_value else None

这将返回最短匹配项,如果没有匹配项,则返回None。

First, halve the string as long as it’s a “2 part” duplicate. This reduces the search space if there are an even number of repeats. Then, working forwards to find the smallest repeating string, check if splitting the full string by increasingly larger sub-string results in only empty values. Only sub-strings up to length // 2 need to be tested since anything over that would have no repeats.

def shortest_repeat(orig_value):
    if not orig_value:
        return None

    value = orig_value

    while True:
        len_half = len(value) // 2
        first_half = value[:len_half]

        if first_half != value[len_half:]:
            break

        value = first_half

    len_value = len(value)
    split = value.split

    for i in (i for i in range(1, len_value // 2) if len_value % i == 0):
        if not any(split(value[:i])):
            return value[:i]

    return value if value != orig_value else None

This returns the shortest match or None if there is no match.


回答 6

O(n)在最坏的情况下,也可以使用前缀功能解决该问题。

请注意,这可能是在一般的情况下慢(UPD:是慢得多),比取决于除数的一些其他的解决方案n,我认为不好的情况下,他们一会,但通常会发现失败越早aaa....aab,那里n - 1 = 2 * 3 * 5 * 7 ... *p_n - 1 a

首先需要计算前缀函数

def prefix_function(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    return pi

那么要么没有答案,要么最短的时间是

k = len(s) - prefix_function(s[-1])

并且您只需要检查是否k != n and n % k == 0(如果k != n and n % k == 0答案为s[:k],则没有答案

您可以在此处检查证明(俄语,但在线翻译可能会解决问题)

def riad(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    k = n - pi[-1]
    return s[:k] if (n != k and n % k == 0) else None

The problem may also be solved in O(n) in worst case with prefix function.

Note, it may be slower in general case(UPD: and is much slower) than other solutions which depend on number of divisors of n, but usually find fails sooner, I think one of bad cases for them will be aaa....aab, where there are n - 1 = 2 * 3 * 5 * 7 ... *p_n - 1 a‘s

First of all you need to calculate prefix function

def prefix_function(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    return pi

then either there’s no answer or the shortest period is

k = len(s) - prefix_function(s[-1])

and you just have to check if k != n and n % k == 0 (if k != n and n % k == 0 then answer is s[:k], else there’s no answer

You may check the proof here (in Russian, but online translator will probably do the trick)

def riad(s):
    n = len(s)
    pi = [0] * n
    for i in xrange(1, n):
        j = pi[i - 1]
        while(j > 0 and s[i] != s[j]):
            j = pi[j - 1]
        if (s[i] == s[j]):
            j += 1
        pi[i] = j;
    k = n - pi[-1]
    return s[:k] if (n != k and n % k == 0) else None

回答 7

这个版本只尝试那些影响字符串长度的候选序列长度。并使用*运算符从候选序列中构建一个全长字符串:

def get_shortest_repeat(string):
    length = len(string)
    for i in range(1, length // 2 + 1):
        if length % i:  # skip non-factors early
            continue

        candidate = string[:i]
        if string == candidate * (length // i):
            return candidate

    return None

感谢TigerhawkT3注意到,length // 2如果不+ 1这样做,将无法匹配abab案件。

This version tries only those candidate sequence lengths that are factors of the string length; and uses the * operator to build a full-length string from the candidate sequence:

def get_shortest_repeat(string):
    length = len(string)
    for i in range(1, length // 2 + 1):
        if length % i:  # skip non-factors early
            continue

        candidate = string[:i]
        if string == candidate * (length // i):
            return candidate

    return None

Thanks to TigerhawkT3 for noticing that length // 2 without + 1 would fail to match the abab case.


回答 8

这是没有正则表达式的直接解决方案。

对于s从零索引开始的,长度为1到的len(s)子字符串,请检查该子字符串substr是否为重复模式。可以通过将substr其自身的ratio时间串联在一起来执行此检查,以使由此形成的字符串的长度等于的长度s。因此ratio=len(s)/len(substr)

当找到第一个这样的子字符串时返回。如果存在,这将提供最小的子字符串。

def check_repeat(s):
    for i in range(1, len(s)):
        substr = s[:i]
        ratio = len(s)/len(substr)
        if substr * ratio == s:
            print 'Repeating on "%s"' % substr
            return
    print 'Non repeating'

>>> check_repeat('254725472547')
Repeating on "2547"
>>> check_repeat('abcdeabcdeabcdeabcde')
Repeating on "abcde"

Here’s a straight forward solution, without regexes.

For substrings of s starting from zeroth index, of lengths 1 through len(s), check if that substring, substr is the repeated pattern. This check can be performed by concatenating substr with itself ratio times, such that the length of the string thus formed is equal to the length of s. Hence ratio=len(s)/len(substr).

Return when first such substring is found. This would provide the smallest possible substring, if one exists.

def check_repeat(s):
    for i in range(1, len(s)):
        substr = s[:i]
        ratio = len(s)/len(substr)
        if substr * ratio == s:
            print 'Repeating on "%s"' % substr
            return
    print 'Non repeating'

>>> check_repeat('254725472547')
Repeating on "2547"
>>> check_repeat('abcdeabcdeabcdeabcde')
Repeating on "abcde"

回答 9

我从八个以上的解决方案开始。一些基于正则表达式(match,findall,split),一些基于字符串切片和测试,而另一些基于字符串方法(find,count,split)。每种代码在代码清晰度,代码大小,速度和内存消耗方面都有好处。当我注意到执行速度被列为重要事项时,我将在此处发布答案,因此我进行了更多测试和改进以得出结论:

def repeating(s):
    size = len(s)
    incr = size % 2 + 1
    for n in xrange(1, size//2+1, incr):
        if size % n == 0:
            if s[:n] * (size//n) == s:
                return s[:n]

该答案似乎与此处的其他一些答案相似,但是它具有一些其他人未使用的速度优化:

  • xrange 在这个应用程序中速度更快,
  • 如果输入字符串是奇数长度,请不要检查任何偶数长度的子字符串,
  • 通过s[:n]直接使用,我们避免在每个循环中创建变量。

我很想看看它在常见硬件的标准测试中如何执行。我相信,在大多数测试中,这将远远超出David Zhang的出色算法,但否则应该很快。

我发现这个问题非常违反直觉。我认为很快的解决方案很慢。看起来很慢的解决方案很快!看起来,使用乘法运算符和字符串比较对Python的字符串创建进行了高度优化。

I started with more than eight solutions to this problem. Some were bases on regex (match, findall, split), some of string slicing and testing, and some with string methods (find, count, split). Each had benefits in code clarity, code size, speed and memory consumption. I was going to post my answer here when I noticed that execution speed was ranked as important, so I did more testing and improvement to arrive at this:

def repeating(s):
    size = len(s)
    incr = size % 2 + 1
    for n in xrange(1, size//2+1, incr):
        if size % n == 0:
            if s[:n] * (size//n) == s:
                return s[:n]

This answer seems similar to a few other answers here, but it has a few speed optimisations others have not used:

  • xrange is a little faster in this application,
  • if an input string is an odd length, do not check any even length substrings,
  • by using s[:n] directly, we avoid creating a variable in each loop.

I would be interested to see how this performs in the standard tests with common hardware. I believe it will be well short of David Zhang’s excellent algorithm in most tests, but should be quite fast otherwise.

I found this problem to be very counter-intuitive. The solutions I thought would be fast were slow. The solutions that looked slow were fast! It seems that Python’s string creation with the multiply operator and string comparisons are highly optimised.


回答 10

此功能运行非常快(经过测试,在超过10万个字符的字符串上,此功能比最快的解决方案快3倍以上,并且重复模式越长,差异越大)。它试图最小化获得答案所需的比较次数:

def repeats(string):
    n = len(string)
    tried = set([])
    best = None
    nums = [i for i in  xrange(2, int(n**0.5) + 1) if n % i == 0]
    nums = [n/i for i in nums if n/i!=i] + list(reversed(nums)) + [1]
    for s in nums:
        if all(t%s for t in tried):
            print 'Trying repeating string of length:', s
            if string[:s]*(n/s)==string:
                best = s
            else:
                tried.add(s)
    if best:
        return string[:best]

请注意,例如对于长度为8的字符串,它仅检查大小为4的片段,并且不必进一步测试,因为长度为1或2的模式将导致重复长度为4的模式:

>>> repeats('12345678')
Trying repeating string of length: 4
None

# for this one we need only 2 checks 
>>> repeats('1234567812345678')
Trying repeating string of length: 8
Trying repeating string of length: 4
'12345678'

This function runs very quickly (tested and it’s over 3 times faster than fastest solution here on strings with over 100k characters and the difference gets bigger the longer the repeating pattern is). It tries to minimise the number of comparisons needed to get the answer:

def repeats(string):
    n = len(string)
    tried = set([])
    best = None
    nums = [i for i in  xrange(2, int(n**0.5) + 1) if n % i == 0]
    nums = [n/i for i in nums if n/i!=i] + list(reversed(nums)) + [1]
    for s in nums:
        if all(t%s for t in tried):
            print 'Trying repeating string of length:', s
            if string[:s]*(n/s)==string:
                best = s
            else:
                tried.add(s)
    if best:
        return string[:best]

Note that for example for string of length 8 it checks only fragment of size 4 and it does not have to test further because pattern of length 1 or 2 would result in repeating pattern of length 4:

>>> repeats('12345678')
Trying repeating string of length: 4
None

# for this one we need only 2 checks 
>>> repeats('1234567812345678')
Trying repeating string of length: 8
Trying repeating string of length: 4
'12345678'

回答 11

在David Zhang的回答中,如果我们有某种循环缓冲区,这将不起作用:principal_period('6210045662100456621004566210045662100456621')由于开始的原因621,我希望将其吐出:00456621

扩展他的解决方案,我们可以使用以下方法:

def principal_period(s):
    for j in range(int(len(s)/2)):
        idx = (s[j:]+s[j:]).find(s[j:], 1, -1)
        if idx != -1:
            # Make sure that the first substring is part of pattern
            if s[:j] == s[j:][:idx][-j:]:
                break

    return None if idx == -1 else s[j:][:idx]

principal_period('6210045662100456621004566210045662100456621')
>>> '00456621'

In David Zhang’s answer if we have some sort of circular buffer this will not work: principal_period('6210045662100456621004566210045662100456621') due to the starting 621, where I would have liked it to spit out: 00456621.

Extending his solution we can use the following:

def principal_period(s):
    for j in range(int(len(s)/2)):
        idx = (s[j:]+s[j:]).find(s[j:], 1, -1)
        if idx != -1:
            # Make sure that the first substring is part of pattern
            if s[:j] == s[j:][:idx][-j:]:
                break

    return None if idx == -1 else s[j:][:idx]

principal_period('6210045662100456621004566210045662100456621')
>>> '00456621'

回答 12

这是python中的代码,用于检查用户给定的主字符串中子字符串的重复

print "Enter a string...."
#mainstring = String given by user
mainstring=raw_input(">")
if(mainstring==''):
    print "Invalid string"
    exit()
#charlist = Character list of mainstring
charlist=list(mainstring)
strarr=''
print "Length of your string :",len(mainstring)
for i in range(0,len(mainstring)):
    strarr=strarr+charlist[i]
    splitlist=mainstring.split(strarr)
    count = 0
    for j in splitlist:
        if j =='':
            count+=1
    if count == len(splitlist):
        break
if count == len(splitlist):
    if count == 2:
        print "No repeating Sub-String found in string %r"%(mainstring)

    else:
        print "Sub-String %r repeats in string %r"%(strarr,mainstring)
else :
    print "No repeating Sub-String found in string %r"%(mainstring)

输入

0045662100456621004566210045662100456621

输出

琴弦长度:40

子字符串’00456621’在字符串’0045662100456621004566210045662100456621’中重复

输入

004608294930875576​​036866359447

输出

字符串长度:30

在字符串’004608294930875576​​576036866359447’中找不到重复的子字符串

Here is the code in python that checks for repetition of sub string in the main string given by the user.

print "Enter a string...."
#mainstring = String given by user
mainstring=raw_input(">")
if(mainstring==''):
    print "Invalid string"
    exit()
#charlist = Character list of mainstring
charlist=list(mainstring)
strarr=''
print "Length of your string :",len(mainstring)
for i in range(0,len(mainstring)):
    strarr=strarr+charlist[i]
    splitlist=mainstring.split(strarr)
    count = 0
    for j in splitlist:
        if j =='':
            count+=1
    if count == len(splitlist):
        break
if count == len(splitlist):
    if count == 2:
        print "No repeating Sub-String found in string %r"%(mainstring)

    else:
        print "Sub-String %r repeats in string %r"%(strarr,mainstring)
else :
    print "No repeating Sub-String found in string %r"%(mainstring)

Input:

0045662100456621004566210045662100456621

Output :

Length of your string : 40

Sub-String ‘00456621’ repeats in string ‘0045662100456621004566210045662100456621’

Input :

004608294930875576036866359447

Output:

Length of your string : 30

No repeating Sub-String found in string ‘004608294930875576036866359447’